2025 AMC 12A Problems/Problem 6: Difference between revisions

Revision as of 02:26, 9 November 2025

The following problem is from both the 2025 AMC 10A #14 and 2025 AMC 12A #6, so both problems redirect to this page.

Problem

Six chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?

$\textbf{(A) } \frac 16 \qquad \textbf{(B) } \frac 15 \qquad \textbf{(C) } \frac 29 \qquad \textbf{(D) } \frac 3{13} \qquad \textbf{(E) } \frac 14$

Solution 1

We first count the number of ways to place $4$ distinct people into $6$ distinct chairs: $6\cdot5\cdot4\cdot3 = 360$ .

We now count how many favorable assignments there are. There are $6$ ways to choose an adjacent pair of chairs for the two students. We can arrange the two students in those two chairs in $2$ ways.

After those two adjacent chairs are taken, there are $4$ chairs left, with $3$ adjacent pairs among them. We choose one of these pairs, arranging the teachers for $3 \cdot 2 = 6$ ways.

There are $6\cdot 2 \cdot 6 = 72$ favorable arrangements.

The probability is therefore $\frac{72}{360} = \boxed{\frac{1}{5}}$

~lprado

Solution 2

We split the problem into cases of where the first teacher and first student sits. The first student sits in a seat with $1$ probability

Case $1$ :

The first teacher sits next to the first student. There's are $2$ ways to do this so this will happen with $\frac{2}{5}$ chance. Now, there is one valid seat for the second student to sit in with probability $\frac{1}{4}$ and and one valid valid seat for the second teacher to sit in with probability $\frac{1}{3}$ . Total probability of this case is $\frac{2}{5}\cdot \frac{1}{4}\cdot \frac{1}{3} = \frac{1}{30}$

Case $2$ :

The first teacher sits on the opposite end of the circle with $1/5$ chance. This means the second student and the second teacher each have $2$ valid spots to sit in, with probability $\frac{2}{4}$ and $\frac{2}{3}$ . Total probability is $\frac{1}{5}\cdot \frac{1}{2}\cdot \frac{2}{3} = \frac{1}{15}$ .

Case $3$ :

The first teacher sits one chair away from the first student with $\frac{2}{5}$ probability because there are $2$ seats that are one chair away.

Case $3.1$ :

The second student sits in between the first teacher and first student with $\frac{1}{4}$ chance. The second teacher only has one valid seat, with $\frac{1}{3}$ chance to sit in it. Total probability is $\frac{1}{4}\cdot \frac{1}{3} = \frac{1}{12}$ .

Case $3.2$ :

The second student doesn't sit in between the teacher and the student with a $\frac{1}{4}$ chance. The second teacher can sit in $2$ valid seats with a $\frac{2}{3}$ chance. Total probability is $\frac{1}{4}\cdot \frac{2}{3} = \frac{1}{6}$ .

Total probability for case $3$ is $\frac{2}{5}\cdot (\frac{1}{12} + \frac{1}{6}) = \frac{1}{10}$ .

Adding up all the cases, you get $\frac{1}{30} + \frac{1}{15} + \frac{1}{10} = \boxed{\frac{1}{5}}$ .

~ Logibyte

Solution 3

We consider how many ways there are to place the students and teachers at random at the table. This is a word arrangement in a circle, and so there are $\frac{(6-1)!}{2! \cdot 2!\cdot 2!}=15$ different ways of doing so, assuming that teachers, students and empty spaces are indistinguishable between themselves.

Now, we wish to find how many favorable arrangments there are, so suppose we fix two adjacent teachers. Then, there are three "spots" (blocks of two seats) two adjacent students could occupy: two on the side of either teacher, and one such that no student is sitting adjacent to a teacher.

Thus, the probability is $\frac{3}{15}=\boxed{\frac{1}{5}}$ .

~e_is_2.71828

Solution 4 (Step-by-step, needs verification)

Because this is in a circle, rotating around doesn't increase or decrease the probability (so pin one seat and the others will not rotate).

Hence, label the spots 1-6 in clockwise. Let student 1 sit in spot 1. Student 2 can therefore sit in either spot 2 or 6, which the probability is 2/5. For explanation purposes, put student 2 in position 2.

Here then requires a case breakdown.

Case 1: Teacher $1$ sits in spot $3$ , hence teacher $2$ needs to sit in spot $4$ to satisfy. This case totals up with $\frac{2}{5}\cdot \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{30}$ .

Case 2: Teacher $1$ sits in spot $4$ , hence teacher $2$ needs to sit in spot $3$ or $5$ to satisfy. This case totals up with $\frac{2}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{15}$ .

Case 3: Teacher $1$ sits in spot $5$ , hence teacher $2$ needs to sit in spot $4$ or $6$ to satisfy. This case totals up with $\frac{2}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{15}$ .

Case 4: Teacher $1$ sits in spot $6$ , hence teacher $2$ needs to sit in spot $5$ to satisfy. This case totals up with $\frac{2}{5} \cdot \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{30}$ .

By adding the case probabilities, we have $(\frac{1}{30})\cdot 2 + (\frac{1}{15}) \cdot 2 = \boxed{\frac{1}{5}}$ .

Note: Feel free to delete if there is a logic error. ~Mitsuihisashi14

~Formatting and $\LaTeX$ by e_is_2.71828

Solution 5 (Quick)

For the two teachers' seatings, there are $6$ ways that they are adjacent out of a combination of $n = \binom{6}{2}$ ways. This probability is $\frac{6}{15} = \frac{2}{5}$ . For the two students' seatings, there are $3$ ways to choose the two seats out of $n = \binom{4}{2}$ ways. This probability is $\frac{3}{6} = \frac{1}{2}$ . Answer is $\frac{2}{5} \cdot \frac{1}{2} = \boxed{\frac{1}{5}}$ .

~Mitsuihisashi14

~Moonlight11 (for latex)

Chinese Video Solution

https://www.bilibili.com/video/BV1FhkUBCEn2/

~metrixgo

Video Solution by Power Solve

https://youtu.be/Vd_kvodRjNQ?si=dmb_X79bIvvymPjF&t=262

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution (Intuitive, Quick Explanation!)

https://youtu.be/3H5VfJTMag8

~ Education, the Study of Everything

2025 AMC 10A (Problems • Answer Key • Resources)
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2025 AMC 12A Problems/Problem 6: Difference between revisions

Revision as of 02:26, 9 November 2025

Contents

Problem

Solution 1